Problem: You just got a free ticket for a boat ride, and you can bring along $3$ friends! Unfortunately, you have $5$ friends who want to come along. How many different groups of friends could you take with you?
Solution: There are $3$ places for your friends on the boat, so let's fill those slots one by one. For the first slot, we have $5$ different choices we can make (because $5$ different friends could be in that slot). Once we've filled the first slot, there are $4$ friends who could fill the second. So far, if we've filled the first two slots, and it seems like there are $5 \cdot 4 = 20 $ different choices we could have made. But that's not quite true. Why? Because if we picked Umaima, then Vanessa, that's the same thing as picking Vanessa, then Umaima. They both get to be on the same boat. So, if we continue filling the slots on our boat, making $5\cdot4\cdot3 = \dfrac{5!}{2!} = 60$ decisions altogether, we've overcounted a bunch of groups. How much have we overcounted? Well, for every group of $3$ , we've counted them as if the order we chose them in matters, when really it doesn't. So, the number of times we've overcounted each group is the number of ways to order $ 3$ things. There are $3! = 6$ ways of ordering $3$ things, so we've counted each group of $3$ friends $6$ times. So, we have to divide the number of ways we could have filled the boat in order by number of times we've overcounted our groups: $ \dfrac{5!}{2!} \cdot \dfrac{1}{3!}$ is the number of groups we can bring on the boat trip. Another way to write this is $ \binom{5}{3} $, or $5$ choose $3$, which is $10$.